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From this picture,when\$ V_{in} = V_{DD}\$,I know \$M_1\$ and \$M_3\$ will turn on,and \$M_2\$ will turn off because their \$ V_{gs} \$ are \$V_{DD}\$ and 0 ,but why will \$M_4\$ turn off,and \$V_B=V_{DD}\$, \$V_{out}=0\$ ?
If i know \$V_{out}=0\$ and \$V_B=V_{DD}\$ first,then of course i know \$M_4\$ will turn off ,but if now i don't know the value of \$V_B\$ and \$V_{out}\$,and neither do \$M_4\$ ,how do i know \$M_4\$ turn off,and \$V_B=V_{DD}\$, \$V_{out}=0\$ ?
As the gate of M4 (which is a PMOS) is \$V_{DD}\$, the only way to make M4 turn on is the take the bulk or drain or source connection of M4 at least \$V_t\$ above \$V_{DD}\$. So we would need \$V_{DD} + V_t\$ to be present somewhere in the circuit.
This can only happen if capacitor \$C_B\$ is first charged (with \$V_A\$ = GND and \$V_B\$ = \$V_{DD}\$) and then discharged by making \$V_A = V_{DD}\$ then there would for a short time be \$2*V_{DD}\$ present at \$V_B\$. But as \$C_B\$ discharges through M5, which is on, see Andy's answer, so the \$V_{GS}\$ of M4 must be zero, hence it is off.
If M3 turns on then M5 has to activate and this has to turn M4 off. If M3 is on and M4 off then Vout is zero.
